If 22700cm of material is available to make a boxwith a square base and an open top, find thedimensions (length, width, and height) of the box thatgive the largest volume of the box. What is themaximum volume of the box?

To solve this problem, we need to use calculus and the method of optimization. Here are the steps:

1. First, let's denote the side length of the square base as x (in cm) and the height of the box as h (in cm). The box is open at the top, so the surface area of the box is given by A = x^2 + 4xh.

2. We know that the total amount of material available is 22700 cm^2, so we have the equation x^2 + 4xh = 22700.

3. We can solve this equation for h to express h in terms of x: h = (22700 - x^2) / (4*x).

4. The volume V of the box is given by V = x^2h. Substituting h from step 3, we get V = x^2 * (22700 - x^2) / (4x) = (22700x - x^3) / 4.

5. To find the maximum volume, we need to take the derivative of V with respect to x, set it equal to zero, and solve for x. The derivative is V' = (22700 - 3x^2) / 4.

6. Setting V' = 0 gives the equation 22700 - 3x^2 = 0. Solving for x, we get x = sqrt(22700 / 3) = 87.18 cm (rounded to two decimal places).

7. Substituting x = 87.18 cm back into the equation from step 3, we get h = (22700 - 87.18^2) / (4*87.18) = 65.38 cm (rounded to two decimal places).

8. So, the dimensions that give the largest volume are length = width = 87.18 cm and height = 65.38 cm.

9. The maximum volume is then V = 87.18^2 * 65.38 = 497,000 cm^3 (rounded to the nearest whole number).